The solution of the equation frac{dy}{dx} = f | vedclass

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(A) Given the differential equation $\frac{dy}{dx} = \frac{1}{x + y + 1}$.Taking the reciprocal,we get $\frac{dx}{dy} = x + y + 1$.Rearranging the ter

Vedclass · Accepted Answer

$x = ce^y - y - 2$

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(A) Given the differential equation $\frac{dy}{dx} = \frac{1}{x + y + 1}$.Taking the reciprocal,we get $\frac{dx}{dy} = x + y + 1$.Rearranging the terms,we have $\frac{dx}{dy} - x = y + 1$.This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = -1$ and $Q(y) = y + 1$.The integrating factor $(I.F.)$ is $e^{\int P(y) dy} = e^{\int -1 dy} = e^{-y}$.The general solution is given by $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + c$.$x e^{-y} = \int (y + 1) e^{-y} dy + c$.Using integration by parts for $\int (y + 1) e^{-y} dy$:Let $u = y + 1$ and $dv = e^{-y} dy$,then $du = dy$ and $v = -e^{-y}$.$\int (y + 1) e^{-y} dy = (y + 1)(-e^{-y}) - \int (-e^{-y}) dy = -(y + 1)e^{-y} - e^{-y} + c = -(y + 2)e^{-y} + c$.Thus,$x e^{-y} = -(y + 2)e^{-y} + c$.Multiplying both sides by $e^y$,we get $x = -(y + 2) + ce^y$,which simplifies to $x = ce^y - y - 2$.

The solution of the equation $\frac{dy}{dx} = \frac{1}{x + y + 1}$ is

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